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Is it possible to convert the 2D Gaussian function in to polar coordinates?

$$\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\exp\big(-({(x-\mu_x)^2+(y-\mu_y)^2})/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y $$

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    $\begingroup$ You would want your pole to be at $(\mu_x, \mu_y)$, so $x = \mu_x + r \cos \theta$ and similarly for $y$. $\endgroup$ – Sammy Black Jun 8 '11 at 5:16
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Hint: There are many symmetries at work here. What if $\mu _x = \mu _y = 0$?

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  • $\begingroup$ No it is not the case $$\mu _x != \mu _y $$ $\endgroup$ – shaikh Jun 8 '11 at 6:44
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    $\begingroup$ @shaikh: you missed the point - you can let them both be 0 without any problem to at least discover what to do. This is completely y symmetric around $\mu_y$ and completely x symmetric around $\mu_x$. I will also mention that you are probably familiar with the fact that $\int e^{- (x - \mu_x)^2} dx = \int e^{- x^2}$. $\endgroup$ – davidlowryduda Jun 8 '11 at 6:51
  • $\begingroup$ but $$ \mu _x, \mu _y $$ are important in my case $\endgroup$ – shaikh Jun 8 '11 at 7:01
  • $\begingroup$ @shaikh: that's fine - then you should probably move the x and y polar coordinates by their respective mu... $\endgroup$ – davidlowryduda Jun 8 '11 at 7:03

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